Greedy vs DP How to Know Which to Apply in DSA Java - Build Both Approaches

Greedy vs DP How to Know Which to Apply

📖 Scenario: Imagine you are organizing a small event and need to decide how to allocate your budget to buy snacks. You want to maximize the number of snacks you can buy without exceeding your budget. Sometimes, a simple quick choice works best (greedy), but other times you need to think carefully about all options (dynamic programming).

🎯 Goal: You will create a Java program that helps decide whether a greedy approach or a dynamic programming approach is better for choosing snacks within a budget. You will practice setting up data, configuring a budget, applying both approaches, and printing the results to compare.

📋 What You'll Learn

Create an array of snack prices

Create a budget variable

Implement a greedy method to pick snacks

Implement a dynamic programming method to pick snacks

Print the results of both methods

💡 Why This Matters

🌍 Real World

Choosing items within a budget is common in shopping, resource allocation, and planning tasks.

💼 Career

Understanding when to use greedy or dynamic programming helps in solving optimization problems efficiently in software development.

1

Create the snack prices array

Create an integer array called snackPrices with these exact values: 2, 3, 4, 5

DSA Java

// Your code here

Hint

Use int[] snackPrices = {2, 3, 4, 5}; to create the array.

2

Set the budget variable

Create an integer variable called budget and set it to 7

DSA Java

int[] snackPrices = {2, 3, 4, 5};
// Your code here

Hint

Use int budget = 7; to set the budget.

3

Implement the greedy snack picker

Write a method called greedySnackPicker that takes snackPrices and budget as parameters and returns the maximum number of snacks you can buy by always picking the cheapest snack first. Use a simple loop and sorting.

DSA Java

int[] snackPrices = {2, 3, 4, 5};
int budget = 7;

// Your code here

Hint

Sort the prices, then pick snacks from cheapest to expensive until budget is reached.

4

Implement the dynamic programming snack picker and print results

Write a method called dpSnackPicker that takes snackPrices and budget as parameters and returns the maximum number of snacks you can buy using dynamic programming. Then, print the results of greedySnackPicker(snackPrices, budget) and dpSnackPicker(snackPrices, budget).

DSA Java

import java.util.Arrays;

int[] snackPrices = {2, 3, 4, 5};
int budget = 7;

public static int greedySnackPicker(int[] snackPrices, int budget) {
    int count = 0;
    int total = 0;
    int[] sortedPrices = snackPrices.clone();
    Arrays.sort(sortedPrices);
    for (int price : sortedPrices) {
        if (total + price <= budget) {
            total += price;
            count++;
        } else {
            break;
        }
    }
    return count;
}

// Your code here

Hint

Use a 2D array to store max snacks for each snack count and budget. Then print both results.

Practice

1. Which of the following best describes when to use a greedy algorithm instead of dynamic programming?

easy

A. When the problem needs to store all intermediate results

B. When the problem requires checking all possible combinations

C. When the problem has overlapping subproblems but no optimal substructure

D. When the problem has the greedy choice property and optimal substructure

Solution

1. Step 1: Understand greedy choice property

   A greedy algorithm works when choosing the best option at each step leads to a global optimum.

2. Step 2: Recognize optimal substructure

   Optimal substructure means the optimal solution can be built from optimal solutions of subproblems.

3. Final Answer:

   When the problem has the greedy choice property and optimal substructure -> Option D

4. Quick Check:

   Greedy = greedy choice + optimal substructure [OK]

Hint: Use greedy if local best leads to global best [OK]

Common Mistakes:

• Confusing overlapping subproblems with greedy applicability
• Using greedy when problem needs full exploration
• Ignoring optimal substructure requirement

2. Which Java syntax correctly declares a 2D array for dynamic programming table initialization?

easy

A. int[][] dp = new int[5];

B. int dp[][] = new int(5,5);

C. int[][] dp = new int[5][5];

D. int dp[5][5] = new int[][];

Solution

1. Step 1: Recall Java 2D array syntax

   Java uses new int[rows][columns] to declare 2D arrays.

2. Step 2: Check each option

   int[][] dp = new int[5][5]; uses correct syntax. Options A, B, C have syntax errors or incomplete declarations.

3. Final Answer:

   int[][] dp = new int[5][5]; -> Option C

4. Quick Check:

   Correct 2D array declaration = int[][] dp = new int[5][5]; [OK]

Hint: Use new int[rows][cols] for 2D arrays in Java [OK]

Common Mistakes:

• Using parentheses instead of brackets
• Missing size in second dimension
• Incorrect array initialization syntax

3. What is the output of the following Java code snippet?

int[] coins = {1, 2, 5};
int amount = 11;
int count = 0;
for (int i = coins.length - 1; i >= 0; i--) {
    count += amount / coins[i];
    amount %= coins[i];
}
System.out.println(count);

medium

A. 3

B. 5

C. 4

D. 6

Solution

1. Step 1: Trace the loop from largest coin to smallest

   Start with amount=11, coins = [1,2,5]. i=2 (coin=5): count += 11/5=2, amount=11%5=1.

2. Step 2: Continue with next coins

   i=1 (coin=2): count += 1/2=0, amount=1%2=1; i=0 (coin=1): count += 1/1=1, amount=1%1=0.

3. Step 3: Calculate total count

   Total count = 2 + 0 + 1 = 3.

4. Final Answer:

   3 -> Option A

5. Quick Check:

   Greedy coin count = 3 [OK]

Hint: Divide and mod with largest coin first [OK]

Common Mistakes:

• Adding counts incorrectly
• Mixing up division and modulo
• Ignoring order of coins

4. The following code attempts to solve the coin change problem using greedy approach. What is the bug?

int[] coins = {1, 3, 4};
int amount = 6;
int count = 0;
for (int i = coins.length - 1; i >= 0; i--) {
    count += amount / coins[i];
    amount %= coins[i];
}
System.out.println(count);

medium

A. Division by zero error

B. Greedy fails because coin set lacks greedy choice property

C. Array index out of bounds error

D. Incorrect initialization of count variable

Solution

1. Step 1: Understand greedy choice property failure

   Coins {1,3,4} do not guarantee greedy choice property for amount 6.

2. Step 2: Trace greedy approach output

   Greedy picks 4 once (count=1), remainder 2; then picks two 1s (count=3). But optimal is two 3s (count=2).

3. Final Answer:

   Greedy fails because coin set lacks greedy choice property -> Option B

4. Quick Check:

   Greedy fails if no greedy choice property [OK]

Hint: Check if greedy choice property holds before using greedy [OK]

Common Mistakes:

• Assuming greedy always works for coin change
• Ignoring optimal substructure requirement
• Confusing runtime errors with logic errors

5. Given a set of tasks with start and end times, which approach is best to find the maximum number of non-overlapping tasks?

class Task {
    int start, end;
    Task(int s, int e) { start = s; end = e; }
}

Choose the correct approach and why.

hard

A. Use greedy by sorting tasks by earliest finish time to select tasks

B. Use dynamic programming to check all subsets of tasks

C. Use greedy by sorting tasks by longest duration first

D. Use dynamic programming to store maximum tasks ending at each time

Solution

1. Step 1: Identify problem type

   This is the activity selection problem, classic greedy example.

2. Step 2: Choose greedy approach

   Sorting by earliest finish time ensures maximum tasks without overlap.

3. Step 3: Compare options

   Use greedy by sorting tasks by earliest finish time to select tasks matches known greedy solution; DP is possible but less efficient here.

4. Final Answer:

   Use greedy by sorting tasks by earliest finish time to select tasks -> Option A

5. Quick Check:

   Activity selection = greedy by finish time [OK]

Hint: Sort tasks by finish time for max non-overlapping tasks [OK]

Common Mistakes:

• Sorting by start time or duration instead of finish time
• Using DP unnecessarily increasing complexity
• Ignoring greedy choice property in scheduling